 As the name of this slide seems to suggest, in lecture 25, we've introduced the notion of a field extension. But it turns out that not all field extensions are created equal. In particular, there is a dichotomy when it comes to field extensions that we are very much just interested in. So suppose we have the field extension E over F, and let alpha be an element in the extension field E. We say that the element alpha is so-called algebraic over F. If there exists some polynomial P of X inside the polynomial ring F of join X, such that alpha is a root of the polynomial P. So P of alpha equals 0. If no such polynomial exists, then we say that alpha is a transcendental element. If every element of the field extension E over F is algebraic, then we say that the extension is an algebraic extension. If you don't have an algebraic extension, then you necessarily have a transcendental extension. Although in a transcendental extension, not every element is transcendental. But if there is one transcendental element in a field extension, then the whole extension is called transcendental. Now, in particular, the most trivial field extension is to extend F over itself. This is in fact an algebraic field extension because if you take any element A inside of the field, notice the polynomial X minus A belongs to F of join X. So A is, of course, a root of the polynomial X minus A. So every element of F is algebraic over itself, but of course, for arbitrary field extensions, that might not be the case. If you take, for example, the field C, the elements square root of 2 and I are algebraic over Q because they have polynomials for which they're the roots of. Like if I take X squared minus 2, that's a rational polynomial whose root is the square root of 2. If you take X squared plus 1, those are, again, that's a rational polynomial whose I is a root of it. So these two elements are examples of algebraic elements. On the other hand, if you take the numbers pi and E, these are transcendental elements. There have been some impressive proofs that show there is no polynomial with rational coefficients whose root is pi or E, although the proofs of such things will go beyond the scope of this lecture series. So we will not include them in this video. I suggest you look it up if you want to see them sometime. So with this in mind here, we have some algebraic element alpha in a field extension E over F. Then I want to define the notion of its minimum polynomial. So if alpha is this algebraic element of the field extension E over F, then there exists a unique monic, irreducible polynomial, P of X over, and P of X belongs to the polynomial ring F of X, where the coefficient here is coming from the base field. This polynomial, which we're going to prove to exist, is referred to as the minimum polynomial of alpha over F. It does depend on what the base field is. And unfortunately, there's no real standard notation for this thing. So when we need to describe it, we can call it mu sub F of alpha here. Mu here is the Greek letter for M. So M is standing for minimum polynomial. F that indicates the base field, because as the base field changes, the minimum polynomial changes, it's dependent upon that. And so the minimum polynomial has the property, of course, that P of alpha is equal to zero. So this minimum polynomial has alpha as a root. And if F of alpha equals zero for some polynomial F of X in F of joint X, that is, if F is another polynomial whose has alpha as a root, then in fact, it's the case that P of X divides F of X. So that's why we call it minimal. This polynomial divides every polynomial that has alpha as a root, when your coefficients come from F here. In particular, the simple extension F of joint alpha is isomorphic to the field F of joint X divided by P of X. This second field right here, this is the field we constructed when we did Kroniker's theorem. And there were some details that were left out. We're basically going to solidify those details right here, right? And then in particular, the degree of the extension F of joint alpha over F is equal to the degree of the minimum polynomial. So in particular, this extension is a finite extension. Like I said, finishing the details of Kroniker's theorem that we omitted earlier. So how are we going to do this? So what we're going to do is we're going to construct for this element, for every algebraic element alpha, we're going to construct the evaluation map, phi sub alpha, which is a map from F of joint X into E. And this is a ring homomorphism from F of X, which is a ring to E, which is a field, which is also a ring. This is a ring homomorphism. It has a kernel, which is an ideal of F of joint X. We will call that kernel I. Now, since F is a field, F of joint X is a Euclidean domain. And every Euclidean domain is a principal ideal domain. So therefore, there exists some polynomial in F of joint X such that I is equal to the principal ideal generated by that polynomial P of X. And so this P of X here is our candidate for the minimum polynomial. It's going to be the element that generates the kernel of the evaluation map associated to alpha here, okay? Now, we can assume that this polynomial is monic because if it's leading, it's leading coefficient can't be zero because otherwise it wouldn't be the leading coefficient. You can divide by that leading coefficient because our coefficients come from a field. That coefficient is just a unit, so you can divide by it and you get an associate. The associates generate the same ideals. In fact, a principal ideal, two principal ideals are equal to each other if and only if their generators are associated to each other. So without the lots of generality, we can assume P of X is a monic polynomial. That's good news. Now, consider the following here. If F alpha equals, and I guess I should also mention here that by construction, we do have that P alpha equals zero, right? Because the kernel is equal to, well, that's what I is. It's the kernel of P alpha. Like so, it's generated by P, which in particular means that P belongs to it. So if P belongs to it, that means that P of alpha is equal to zero. So that also needs to be mentioned. So imagine we have some other polynomial, F, which evaluated at alpha gives you zero. Well, that means that F of X belongs to the kernel of the evaluation map, which is I, that means F of X belongs to I. But since I is the principal ideal generated by P of X, that then means that F of X belongs to the principal ideal only if the generator divides P of X in that situation. So P of X divides F of X, P of alpha is equal to zero, like we mentioned. And so that gives us a lot of the statements we've already mentioned, okay? What else? P of X, let's say that we factored it in some way. If P of X equals F of X times G of X, where this is not necessarily the same F anymore. And then if you evaluate it at P, evaluate P of X at alpha, that means you're gonna evaluate these other ones likewise, F of alpha times G of alpha. Since P of alpha equals zero, that means this product equals zero. Now be aware that P of alpha, this is not a polynomial, this is the evaluation. This lives inside of E, E is a field, in particular E is a domain. So by the zero product property, either F of alpha equals zero or G of alpha equals zero. And just for the loss of generality, assume the first one. Since F of alpha equals zero, that means F of X belongs to the ideal. Therefore, again, P of X divides F of X, which then tells you that F of X must be an associate of P of X. And as this is an arbitrary factorization of P, this tells, since every factorization uses an associate, this tells you that P of X is an irreducible polynomial. So now we've constructed our monic irreducible minimum polynomial. And it's unique up to, of course, the fact it's monic. It's unique up to association. And if you choose the leading coefficient to be monic, the leading coefficient to be one, then there's only one associate that'll do exactly that. Now since P of X is an irreducible polynomial, the principle ideal generated by P of X, I, is a maximum ideal because we're in a principle ideal domain. And if you mod out a ring, a commutative ring by a maximum ideal, this gives you a field and that field has a root for P of X. And so this, in fact, gives us an isomorphism between, this gives us isomorphism between F of X mod out P with F here, because you're just gonna use the first isomorphism theorem, right? So we have this map, F of joint X that maps over to E. Over here, this is the evaluation map, like so. In which case then, if we mod out by the kernel of this thing, F of joint X mod out by its kernel, kernel of phi alpha. And the first isomorphism theorem shows that these two rings are isomorphic, thus giving us what we meant before. So this is what I was alluding to earlier when I said simple extensions are the extensions fields that we get from Kronecker's theorem. This is the type of field we constructed in Kronecker's theorem. That's the same thing as this, as this simple extension. All right, so the last thing we wanna prove about the minimum polynomial right now has to do with the degree. So we want to show that F of joint alpha over F, which remember by definition, this is the dimension as an F vector space of the vector space F of joint alpha. So we wanna show that this is equal to the degree of this irreducible polynomial. This minimum polynomial, that's what we have to prove now. So let's suppose that the polynomial P of X looks like A0 plus A1X plus A2X squared. All the way up to AnX to the n. What we saw earlier, as we've talked about these things in previous lectures, that if you view F of X as a F vector space, it has a basis as 1XX squared X cubed all the way up. In particular, it's a spanning set, that's all I really care about. Now if you take a ring homomorphism in this situation, you have F of joint X over to E here. This is a ring homomorphism, but in fact, it's even better than that. I mean, maybe not better, but I should also mention this is gonna be a linear transformation. A linear transformation, remember, is going to be a vector space homomorphism. So this function is gonna preserve addition. You get that from the ring homomorphism. It's also gonna preserve the scalars, which in this case is F. Because F over X, F of X is a F dimensional vector space. E as a field extension is an F dimensional vector space. So this evaluation map is in fact a linear transformation. So linear transformation send spanning sets to spanning sets. Now the image of an independent set under linear transformation is not necessarily linear independent. So we can't argue that this is a basis, but this is in fact a spanning set. That's the thing I want to observe right here. We've used this fact previously without being explicit about it. So if you take one alpha, alpha squared, alpha cubed, alpha to the 5,280th. But take all of those alphas, that gives us a spanning set for F of joint alpha, okay? Now, since we have this polynomial relation, if I were to plug in alpha into this situation, you get alphas for each of these X's, this equals zero. If I moved A n X to the end to the other side and divided by A n, the leading coefficient, which is not zero, we get the following observation right here. Alpha to the n equals negative one over A n times A zero plus A one alpha plus A two alpha squared all the way up to A n minus one times alpha to the n minus one, okay? And so this shows us that alpha to the n can be written as a linear combination of one alpha, alpha squared, alpha cubed all up to alpha n minus one. And then by induction, every larger power of alpha can also do that. So alpha n plus one will give you that because if you take alpha to the n plus one, that's equal to alpha times alpha to the n, alpha to the n looks like this. You distribute that alpha to each of those. Each of these will get larger by a power of alpha. One of them will turn into alpha to the n. You use this formula again to reduce it. And therefore, your largest power of alpha is going to be alpha n minus one. You can do this by induction and show that alpha to the capital N always belongs to the span of one alpha, alpha squared, all the way up to alpha to the n minus one. So this then gives us an even smaller spanning set, like so we're going to call this thing B. And so this gives us a spanning set for F the joint alpha. Now, is it an independent spanning set? That is, is it a basis? That's what we want to argue right now. I claim that this set B is an independent set, hence it's a basis. Now, if it were dependent, that means there would exist some number k such that the expression B0 plus B1 alpha plus B2 alpha squared all the way up to Bk, alpha k equals zero, right? If it was dependent, then some combination of these things, it could be, it could be that k equals n minus one. All that we know about k here is that k is less than or equal to n minus one. Maybe it uses all of them, maybe it uses fewer, it doesn't really matter, okay? In particular, k is less than n because it's at most n minus one. Now, if you look at this right here, we can introduce a polynomial g of x, which looks like B0 plus B1x all the way up to Bk, xk. And this would, of course, be a polynomial inside of F of joint x because the coefficients of the B's are elements that belong to the base field F here. And so notice that, oh, g, when evaluated at alpha, gives you zero. Okay, that polynomial then belongs to the ideal i. But i is generated by P of x, right? So how can that be? We have a polynomial g whose degree is k. We have another polynomial P whose degree is n, n is greater than k. But since g belongs to this ideal, that would suggest that P doesn't have the minimum degree inside of the principal ideal, that's a contradiction. You can't have that P divides g of x because g has a smaller degree than P. That's not how the additive norm of degrees works. We get a contradiction. The contradiction was that we assumed B was dependent, and therefore, it has to be independent. So since B is a basis for this field extension, the dimension of F, a joint alpha over F, will have the same cardinality as B, but B contains exactly n elements. It's not n minus 1 because you have 1, 2, 2, 3, 4, all the way up to n minus 1, but you also have this one right here. There's n elements in there. And this then proves the existence and uniqueness of our minimum polynomials. That brings us to the end of lecture 25 about field extensions. Thanks for watching. If you've learned anything about the minimum polynomial or about algebraic extensions, field extensions, like these videos, subscribe to the channel to see more videos like this in the future, and please post your questions in the comments below. If you have any, I'll be glad to answer them.